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2001/10/25 Sheng-Feng Ho 1
Efficient and Scalable On-Demand Data Streaming Using UEP Codes
Lihao XuWashington University in St. Louis
ACM Multimedia 2001 Sept. 30 – Oct. 5, 2001
2001/10/25 Sheng-Feng Ho 2
OutlineIntroductionUnequal Error Protection Codes UEP Codes Example Theorem
The SchemeResource Consumption Network Bandwidth Client’s Buffer Space Client’s Network Bandwidth
Additional Initial Playout DelayConclusion
2001/10/25 Sheng-Feng Ho 3
Introduction
Streaming Method Unicast : Point-to-point
Resource consumption is proportional to the number of requests.
Multicast : Pyramid or Skyscraper The lowest resource consumption with no
initial data playout delay is proportional to logarithm of the average request arrival rate.
The reliability : Automatic-Repeat-Request 、 Forward Error Correction
2001/10/25 Sheng-Feng Ho 4
Unequal Error Protection Codes
(N,K) block code encodes an original message of K symbols into a codeword of N data symbols of the same size.A data symbol is a general data unit of certain size : a bit, a byte, a packet or a frame.Original K data symbols can be recovered from any M data symbols of its codeword. (M≧K)
2001/10/25 Sheng-Feng Ho 5
Unequal Error Protection Codes
For a UEP code, certain data symbols of its codeword are protected against a greater number of errors than others.For a message m with n data symbols, if the error protection degree of its i-th symbol is Li (i≦1 ≦n), and it is encoded with a UEP code C of N symbols, then any Li symbols of its codeword are sufficient to retrieve the i-th symbol in the original message m.
2001/10/25 Sheng-Feng Ho 6
UEP Codes Example (1)
Message m has 3 symbols of equal size : a, b and c. (m=abc) Partition symbol a into 6 sub-symbols of equal size :
a=a1..a6, b into 9 sub-symbols : b=b1..b9 c into 6 sub-symbols : c1..c6
2001/10/25 Sheng-Feng Ho 7
UEP Codes Example (2)Apply the (6,2) B-Code on a to get a codeword of a : A=A1..A6 (Ai is ½ size of a) A1 = a1, a2+a3, a4+a6, A2 = a2, a3+a4, a5+a1, A3 = a3, a4+a5, a6+a2, A4 = a4, a5+a6, a1+a3, A5 = a5, a6+a1, a2+a4, A6 = a6, a1+a2, a3+a5, and + is the simple bit-wise binary exclusive
or (XOR) operations
2001/10/25 Sheng-Feng Ho 8
UEP Codes Example (3)
Apply a modified (6,3) RS-Code on b to get a codeword of b : B=B1..B6 (Bi is 1/3 size of b) B1 = b1, b2, b3, B2 = b1+b3+b4, b2+b4+b5, b2+b3+b6, B3 = b6+b7, b4+b6+b8, b5+b9, B4 = b3+b6+b7, b1+b4+b7+b8, b2+b5+b9 B5 = b4+b6+b9, b4+b7, ? B6 = b7, b8, b9 “+” is the XOR operation
Let Ci = ci for i = 1 to 6, thus each Ci is 1/6 size of c.
2001/10/25 Sheng-Feng Ho 9
UEP Codes Example (4)
Construct a UEP codeword of the original message m : U = U1..U6, where Ui = AiBiCi for i = 1 to 6.
The protection degrees of the original data symbols a, b and c are La=2, Lb=3 and Lc=6 respectively.
1/La+1/Lb+1/Lc=1
2001/10/25 Sheng-Feng Ho 10
UEP Codes Example (5)
Original Messagem=abc
UEP CodewordU=U1U2U3
U1=A1B1C1,U2=..
Server Multicast Client ReceiveNetwork Transmit
UEP CodewordU=U1U2U3
a=a1..a6,b=b1..
Original Messagem=abc
2001/10/25 Sheng-Feng Ho 11
Theorem
For a message m with n symbols, if there exists a UEP code such that the error protection degree of the i-th symbol in the original message m is Li (i≦1 ≦n), then
1/11
≦
n
iLi
2001/10/25 Sheng-Feng Ho 12
The Scheme (1)
Encoding the original data of n symbols into a UEP codeword of N symbols.Multicast the UEP codeword in a cyclic fashion.Once the number of data symbols in user’s buffer space reaches Li, the user retrieves the i-th symbol of the original data stream and plays it out.
2001/10/25 Sheng-Feng Ho 13
The Scheme (2)
B : normal playout unicast network bandwidthd : the initial playout number of original data symbolsThe peak network bandwidth needed : rpeakB, where rpeak = max(Li/(i+d))The average network bandwidth is raveB, where raveB = Ln/(n+d)
2001/10/25 Sheng-Feng Ho 14
The Scheme (3)
d 0 1 30 60R 12.859 8.87 5.484 4.796C 36.788% 36.790% 36.941% 37.095%
Resource Consumption vs. Initial Playout Delay for Multicasting a 2-hour Video.
d : initial playout delay in seconds
R : normalized backbone network bandwidth
C : normalized client buffer space needed
2001/10/25 Sheng-Feng Ho 15
Resource Consumption (1)
The peak network bandwidth needed :rpeakB, where rpeak = max(Li/(i+d))
Minimize rpeak, set Li/(i+d) = R for all i’s
R=Hn+d-Hd≒αlog2(1+n/d), where ½≦α≦1
12
1,log1/1
1/1
21
1
whereliH
Li
m
im
n
i≦
R≒αlog2(1+n/d), where ½≦α≦1
2001/10/25 Sheng-Feng Ho 16
Resource Consumption (2)
Let Si be the buffer space needed between the retrieval of the i-th and the (i+1)th original data symbols, then
Client’s Buffer Space :n
diHHS didn
ni
)1)((max
10
)(,1,)1
1(1
11 diRlniil
lnS i
i
ji
ji
n
diHHS didn
ni
)1)((max
10
2001/10/25 Sheng-Feng Ho 17
Resource Consumption (3)The normalized network bandwidth a client needs to consume between the retrieval of the i-th and (i+1)th original data symbols is
10, niHHC didni
)(),1
1(1
diRll
RC i
i
jj
i
10, niHHC didni
2001/10/25 Sheng-Feng Ho 18
Additional Initial Playout Delay
d : additional initial playout delay : normalized backbone network bandwidthW : client’s normalized incoming network badnwidth
ddn HHdR )(
)(,
)(),(,/)()(),(
dRWd
dRWdRWnWdRdndWD
))((,))(
11(),( dndRll
dRwddWD nn